Question

In this reaction, a gaseous system with a volume of 3.00 L has a rate of formation of NOCl of 0.0120 M s-1 . The volume of the system is decreased to 1.00 L with no change of temperature. What is the new rate of formation of NOCl?

Answer 1

Answer:

x = 0.324 M s⁻¹

Explanation:

Equation for the reaction can be represented as:

 2 NO(g)   + Cl₂ (g)      ⇄     2NOCl (g)

Rate = K [NO]² [Cl₂]

Concentration = $\frac{numbers of mole (n)}{volume (v)}$

from the question; their number of moles are constant since the species are quite alike.

As such; if Concentration varies inversely proportional to the volume;

we have: Concentration ∝ $\frac{1}{v}$

Concentration = $\frac{1}{v}$

Similarly; the Rate can now be expressed as:

Rate = K [NO]² [Cl₂]

Rate = $(\frac{1}{v}) ^2$ $(\frac{1}{v} )$

Rate = $\frac{1}{v^3}$

We were also told that the in the reaction, the gaseous system has an initial volume of 3.00 L and rate of formation of 0.0120 Ms⁻¹

So we can have:

0.0120 = $\frac{1}{3^3}$

0.0120 = $\frac{1}{27}$   -----Equation (1)

Now; the new rate of formation when the  volume of the system decreased to 1.00 L can now be calculated as:

x = $(\frac{1}{1})^3$

x = 1             ------- Equation (2)

Dividing equation (2) with equation (1); we have:

$\frac{0.0210}{x}$ = $\frac{\frac{1}{27} }{1}$

$\frac{0.0210}{x}$ = $\frac{1}{27}$

x = 0.0120 × 27

x = 0.324 M s⁻¹

∴  the new rate of formation of NOCl = 0.324 M s⁻¹